16_Ch 16 College Physics ProblemCH16 Electrical Energy and Capacitance

16_Ch 16 College Physics ProblemCH16 Electrical Energy and Capacitance

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Unformatted text preview: 50 CHAPTER 16 (b) Q1 =C1 ( ∆V ) = ( 0.050 µ F ) ( 400 V ) = 20.0 µ C Q2 =C2 ( ∆V ) = ( 0.100 µ F ) ( 400 V ) = 40.0 µ C 16.30 (a) For parallel connection, Ceq =C1 + C2 + C3 = ( 5.00 + 4.00 + 9.00 ) µ F = 18.0 µ F (b) For series connection, 1 1 1 1 = + + Ceq C1 C2 C3 1 1 1 1 = + + , giving Ceq = 1.78 µ F Ceq 5.00 µ F 4.00 µ F 9.00 µ F 16.31 (a) Using the rules for combining capacitors in series and in parallel, the circuit is reduced in steps as shown below. The equivalent capacitor is shown to be a 2 .00 µ F capacitor. 4.00 mF 3.00 mF b a 6.00 mF 3.00 mF a c b c 2.00 mF a c 2.00 mF 12.0 V 12.0 V 12.0 V Figure 1 Figure 2 Figure 3 ...
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